Corrected exercises: series with sin, cos, exp and ln terms, 201-SN3-RE (201-NYB)

The ratio is $\dfrac{e^{n+1}}{(n+1)!}\cdot\dfrac{n!}{e^{n}}=\dfrac{e}{n+1}\to 0<1$: the series converges. Its sum is immediate once you recognize the exponential series $\sum_{n\geq 0}\frac{x^{n}}{n!}=e^{x}$ evaluated at $x=e$: the sum is $e^{e}\approx 15.154$.

$\dfrac{a_{n+1}}{a_{n}}=\dfrac{2^{n+1}(n+1)!}{(n+1)^{n+1}}\cdot\dfrac{n^{n}}{2^{n}n!}=2\cdot\dfrac{(n+1)\,n^{n}}{(n+1)^{n+1}}=2\left(\dfrac{n}{n+1}\right)^{n}=\dfrac{2}{\left(1+\frac{1}{n}\right)^{n}}$. Since $\left(1+\frac{1}{n}\right)^{n}\to e$, the ratio tends to $\dfrac{2}{e}\approx 0.736<1$: the series CONVERGES.

The computation is identical and gives $\dfrac{3}{e}\approx 1.104>1$: the series DIVERGES. The threshold is therefore $k=e$: the series $\sum\frac{k^{n}n!}{n^{n}}$ converges for $k<e$ and diverges for $k>e$. Two series that look almost identical fall on opposite sides because $2<e<3$. This is the typical question that penalizes those who answer by guesswork.

$\dfrac{a_{n+1}}{a_{n}}=\dfrac{(n+1)^{2}}{2^{n+1}}\cdot\dfrac{2^{n}}{n^{2}}=\dfrac{1}{2}\left(\dfrac{n+1}{n}\right)^{2}\to\dfrac{1}{2}<1$: convergence. With the given formula at $x=\frac{1}{2}$: $\dfrac{\frac{1}{2}\left(\frac{3}{2}\right)}{\left(\frac{1}{2}\right)^{3}}=\dfrac{\frac{3}{4}}{\frac{1}{8}}=6$. The sum is exactly 6.

The ratio is $\dfrac{\ln(n+1)}{n+1}\cdot\dfrac{n}{\ln n}\to 1$: the ratio test is INCONCLUSIVE. This is the general situation for series whose terms decrease only polynomially, and on them this test is blind. You conclude another way: for $n\geq 3$, $\ln n>1$ so $\dfrac{\ln n}{n}>\dfrac{1}{n}$, and the harmonic series diverges. By the Direct Comparison Test, the series DIVERGES.

The sequence $\frac{1}{n^{2}}$ is positive, decreasing and tends to 0: the Alternating Series Test applies and the series converges. The remainder after $N$ terms satisfies $|R_{N}|\leq\dfrac{1}{(N+1)^{2}}$. We want $\dfrac{1}{(N+1)^{2}}<10^{-3}$, that is $(N+1)^{2}>1000$, so $N+1>31.6$ and $N\geq 31$. Check: with 31 terms, the actual gap with $\frac{\pi^{2}}{12}\approx 0.8224670$ is $5.0\times 10^{-4}$, indeed below the bound.

Convergence: $\sin\frac{1}{n}$ is positive for $n\geq 1$, decreasing (because $\frac{1}{n}$ decreases in $(0,1]$, where the sine is increasing) and tends to $\sin 0=0$. The test applies, so the series converges. Absolute convergence: $\sin\frac{1}{n}\sim\frac{1}{n}$, so by the Limit Comparison Test $\sum\left|(-1)^{n}\sin\frac{1}{n}\right|$ behaves like the harmonic series, which diverges. The series is therefore CONDITIONALLY CONVERGENT.

$\dfrac{n}{n^{2}+1}$ is positive, tends to 0, and decreases from $n=1$ on. The alternating series converges. In absolute value, $\dfrac{n}{n^{2}+1}\sim\dfrac{1}{n}$, so the series of absolute values diverges: once again, CONDITIONALLY CONVERGENT. The trap would be to conclude absolute convergence on seeing an $n^{2}$ in the denominator, without noticing the $n$ in the numerator.

Here $\left|a_{n}\right|=1+\frac{1}{n}\to 1\neq 0$. The hypothesis of a zero limit fails, so the Alternating Series Test does not apply. Better still: the general term does not tend to 0, so by the Test for Divergence the series DIVERGES outright. Alternating signs change nothing: they never rescue a series whose terms do not vanish at infinity.

Absolutely convergent means that $\sum|a_{n}|$ converges, which implies that $\sum a_{n}$ converges. Conditionally convergent means that $\sum a_{n}$ converges while $\sum|a_{n}|$ diverges. A major practical consequence: in an absolutely convergent series you can reorder the terms freely without changing the sum, whereas in a conditionally convergent series, rearranging the terms can produce ANY sum at all (Riemann's rearrangement theorem). Absolute convergence is therefore the only kind that allows you to handle the sum like a finite sum.

Since $\frac{1}{n^{2}}\to 0$ and $\sin u\sim u$, we have $\sin\frac{1}{n^{2}}\sim\frac{1}{n^{2}}$. The terms are positive and the Limit Comparison Test applies: same behaviour as $\sum\frac{1}{n^{2}}$, a p-series with $p=2>1$. The series CONVERGES.

$1-\cos u\sim\dfrac{u^{2}}{2}$, so $1-\cos\frac{1}{n}\sim\dfrac{1}{2n^{2}}$. Same behaviour as $\sum\frac{1}{n^{2}}$: the series CONVERGES. The factor $\frac{1}{2}$ changes nothing; a nonzero constant never affects whether a series converges.

$\tan u\sim u$, so $\tan\frac{1}{n}\sim\frac{1}{n}$. Same behaviour as the harmonic series: the series DIVERGES. Compare with question a: the square in the denominator makes all the difference, while the sine and the tangent play the same role.

You have to see through the disguise: $\cos(n\pi)=(-1)^{n}$ for every integer $n$. The series is therefore $\sum\frac{(-1)^{n}}{n}$, the alternating harmonic series. It converges by the Alternating Series Test, but not absolutely: it is CONDITIONALLY CONVERGENT, and its sum is $-\ln 2\approx-0.693$. Treating it as a complicated series with terms of arbitrary sign, without recognizing the $(-1)^{n}$, throws away the whole point of the question.

No equivalent is sought here, because $|\sin n|$ oscillates without a limit. Bound it instead: $0\leq\dfrac{|\sin n|}{n^{2}}\leq\dfrac{1}{n^{2}}$, and $\sum\frac{1}{n^{2}}$ converges. By the Direct Comparison Test, the series CONVERGES (absolutely, in fact). This is the reflex to have whenever a sine or cosine of argument $n$ appears: it has no equivalent, but it is bounded, which is almost always enough.

Ratio test: $\dfrac{(n+1)^{2}e^{-(n+1)}}{n^{2}e^{-n}}=\dfrac{1}{e}\left(\dfrac{n+1}{n}\right)^{2}\to\dfrac{1}{e}\approx 0.368<1$: the series CONVERGES. With the formula at $x=e^{-1}$: $\dfrac{e^{-1}(1+e^{-1})}{(1-e^{-1})^{3}}\approx 1.9923$.

The ratio test gives $e^{-\left(\sqrt{n+1}-\sqrt{n}\right)}$, and since $\sqrt{n+1}-\sqrt{n}=\dfrac{1}{\sqrt{n+1}+\sqrt{n}}\to 0$, the ratio tends to $e^{0}=1$: the test is inconclusive. Use the Integral Test, the function $x\mapsto e^{-\sqrt{x}}$ being positive, continuous and decreasing. With $u=\sqrt{x}$, an antiderivative is $-2\left(\sqrt{x}+1\right)e^{-\sqrt{x}}$, whose limit at $+\infty$ is 0. The integral $\int_{1}^{\infty}e^{-\sqrt{x}}dx$ is therefore finite and the series CONVERGES. An exponential in $-\sqrt{n}$ decreases much more slowly than one in $-n$, but still fast enough to beat any power.

$e^{u}-1\sim u$ gives $e^{1/n}-1\sim\dfrac{1}{n}$. By the Limit Comparison Test with the harmonic series, the series DIVERGES. This is the contrast with question a: it is not the presence of an exponential that decides, it is its argument. Here the argument tends to 0, so the exponential tends to 1 and crushes nothing at all.

For large $n$, $e^{n}-n\sim e^{n}$ because $n$ is negligible compared with $e^{n}$. Hence $\dfrac{1}{e^{n}-n}\sim e^{-n}$, the general term of a geometric series with ratio $\frac{1}{e}<1$, which converges. By the Limit Comparison Test, the series CONVERGES. You must still note that $e^{n}-n>0$ for every $n\geq 1$; otherwise the comparison between series of positive terms would be illegitimate.

Ratio test: $\dfrac{(n+1)e^{-(n+1)^{2}}}{n\,e^{-n^{2}}}=\dfrac{n+1}{n}\,e^{-(2n+1)}\to 0<1$. The series CONVERGES, and very quickly: its terms decrease like $e^{-n^{2}}$, so a few terms are enough to obtain the sum to about ten decimal places.

You cannot compare directly with $\frac{1}{n^{2}}$, because $\frac{\ln n}{n^{2}}$ is larger than it. Slip in an intermediate power: $\ln n$ is negligible compared with $n^{1/2}$, so from some index on $\dfrac{\ln n}{n^{2}}\leq\dfrac{n^{1/2}}{n^{2}}=\dfrac{1}{n^{3/2}}$, the term of a convergent p-series ($p=\frac{3}{2}>1$). The series CONVERGES. The Integral Test gives the same conclusion.

For $n\geq 3$ we have $\ln n>1$, so $\dfrac{\ln n}{n}>\dfrac{1}{n}$. The harmonic series diverges, so by the Direct Comparison Test the series DIVERGES. Here the logarithm, however slow, makes a divergence that was already there even worse.

Write $\ln\left(1+\frac{1}{n}\right)=\ln\dfrac{n+1}{n}=\ln(n+1)-\ln n$: the sum telescopes. Hence $S_{N}=\ln(N+1)-\ln 1=\ln(N+1)$, which tends to $+\infty$: the series DIVERGES. The equivalent $\ln\left(1+\frac{1}{n}\right)\sim\frac{1}{n}$ gives the same result, but the telescoping computation is stronger: it gives the rate of divergence, like $\ln N$, which no comparison provides.

The root test is the right one here, because the term is an $n$th power: $\sqrt[n]{a_{n}}=\dfrac{1}{\ln n}\to 0<1$. The series CONVERGES. The ratio test would be much more painful because of the ratio $\left(\frac{\ln n}{\ln(n+1)}\right)^{n}$. Practical rule: as soon as an $n$th power appears as a block, take the root, not the ratio.

Integral Test, with the substitution $u=\ln x$: $\int_{2}^{\infty}\frac{dx}{x(\ln x)^{3/2}}=\int_{\ln 2}^{\infty}\frac{du}{u^{3/2}}$, which converges because $\frac{3}{2}>1$. The series CONVERGES. The general rule is that $\sum\frac{1}{n(\ln n)^{p}}$ converges if and only if $p>1$: the logarithm in the denominator replays the p-series test exactly, one level further.

If $\alpha>1$: we have $\dfrac{1}{n^{\alpha}\ln n}\leq\dfrac{1}{n^{\alpha}}$ as soon as $\ln n\geq 1$, and the p-series converges: CONVERGES. If $\alpha<1$: choose $\beta$ between $\alpha$ and 1; since $\ln n$ is negligible compared with $n^{\beta-\alpha}$, we have $\dfrac{1}{n^{\alpha}\ln n}\geq\dfrac{1}{n^{\beta}}$ from some index on, and $\sum\frac{1}{n^{\beta}}$ diverges: DIVERGES. Boundary case $\alpha=1$: the series becomes $\sum\frac{1}{n\ln n}$, for which the Integral Test gives $\int\frac{du}{u}=\ln u\to\infty$: DIVERGES. Conclusion: converges if and only if $\alpha>1$.

Since $\alpha>0$, $\frac{1}{n^{\alpha}}\to 0$ and $\sin\frac{1}{n^{\alpha}}\sim\frac{1}{n^{\alpha}}$. By the Limit Comparison Test with the p-series: converges if and only if $\alpha>1$. Note that the hypothesis $\alpha>0$ is essential for the equivalent to hold; if $\alpha\leq 0$, the term does not tend to 0 and the series diverges outright.

Ratio test: $\dfrac{(n+1)^{\alpha}e^{-(n+1)}}{n^{\alpha}e^{-n}}=\left(\dfrac{n+1}{n}\right)^{\alpha}e^{-1}\to\dfrac{1}{e}<1$ for EVERY real $\alpha$. The series converges whatever the value of $\alpha$. This is the perfect illustration of the fact that the exponential crushes any power, however large.

Integral Test with $u=\ln x$: the integral becomes $\int_{\ln 2}^{\infty}\frac{du}{u^{\alpha}}$, which converges if and only if $\alpha>1$. The case $\alpha=1$ diverges, which matches the boundary found in a. Remember that $\sum\frac{1}{n(\ln n)^{\alpha}}$ behaves exactly like $\sum\frac{1}{n^{\alpha}}$, with the logarithm playing the role of the variable.

$1-\cos u\sim\frac{u^{2}}{2}$ gives $1-\cos\frac{1}{n^{\alpha}}\sim\dfrac{1}{2n^{2\alpha}}$. By the Limit Comparison Test, the series converges if and only if $2\alpha>1$, that is $\alpha>\frac{1}{2}$. The threshold is therefore at $\frac{1}{2}$ and not at 1: it is the square in the equivalent that moves the boundary, and that is precisely what the question tests.

$\frac{1}{\sqrt{n}}$ decreases to 0: the alternating series converges. But $\sum\frac{1}{\sqrt{n}}$ is a p-series with $p=\frac{1}{2}\leq 1$, so it diverges. CONDITIONALLY CONVERGENT.

In absolute value, $\sum\frac{1}{n^{3}}$ converges ($p=3>1$). ABSOLUTELY CONVERGENT, and therefore convergent. Here studying the alternating series is not even needed: absolute convergence is enough.

The series is not alternating in the strict sense, since the sign of $\sin n$ is irregular. Go straight to the absolute values: $\left|\frac{\sin n}{n^{2}}\right|\leq\frac{1}{n^{2}}$, a convergent series. ABSOLUTELY CONVERGENT. It is the only workable route, and it works because the sine is bounded.

The sequence $\frac{\ln n}{n}$ tends to 0 and decreases for $n\geq 3$ (its derivative $\frac{1-\ln x}{x^{2}}$ is negative as soon as $x>e$): the alternating series converges. But $\sum\frac{\ln n}{n}$ diverges, as shown in exercise 5. CONDITIONALLY CONVERGENT. Careful: the decrease only holds from index 3 on, which must be stated; the first terms do not change whether the series converges.

$\left|a_{n}\right|=\frac{n}{n+1}\to 1\neq 0$. The general term does not tend to 0: DIVERGENT, outright. Neither of the other two answers is possible, since convergence requires at the very least that the term vanish at infinity.

In absolute value, $\sum\frac{1}{n!}$ converges (the ratio test gives $\frac{1}{n+1}\to 0$). ABSOLUTELY CONVERGENT. For the sum, recognize the exponential series at $x=-1$: $\sum_{n\geq 0}\frac{(-1)^{n}}{n!}=e^{-1}$. Removing the term $n=0$, which equals 1, the requested sum is $e^{-1}-1\approx-0.632$.

For $\sum\frac{1}{n}$: $\dfrac{a_{n+1}}{a_{n}}=\dfrac{n}{n+1}\to 1$. For $\sum\frac{1}{n^{2}}$: $\dfrac{n^{2}}{(n+1)^{2}}\to 1$ as well.

Yet the first diverges and the second converges. Two series with opposite behaviours therefore give the same limit $L=1$: this value carries no information, and the test is genuinely undecided, not merely silent out of caution. Writing “$L=1$, so the series diverges” is a serious and frequent mistake.

No, the root test does no better. $\sqrt[n]{\frac{1}{n}}=n^{-1/n}=e^{-\frac{\ln n}{n}}\to e^{0}=1$, and likewise $\sqrt[n]{\frac{1}{n^{2}}}=e^{-\frac{2\ln n}{n}}\to 1$. The two tests fail together, and this is no accident: both are insensitive to polynomial decrease, because they only ever compare the series with geometric series.

The ratio and root tests are made for series whose terms decrease GEOMETRICALLY, that is, terms containing factorials, $k^{n}$ or $n$th powers. Avoid them from the outset as soon as the term is a rational function of $n$, possibly multiplied by a logarithm: in that case the ratio always tends to 1. Switch instead to the p-series test, the Limit Comparison Test or the Integral Test.

$\sum\frac{n!}{n^{n}}$: a factorial and an $n$th power, so the ratio test; the ratio tends to $\frac{1}{e}<1$, CONVERGES. $\sum\frac{1}{n\sqrt{n}}=\sum\frac{1}{n^{3/2}}$: a pure power, so the p-series test directly; $p=\frac{3}{2}>1$, CONVERGES. $\sum\left(\frac{n}{2n+1}\right)^{n}$: an $n$th power as a block, so the root test; the root is $\frac{n}{2n+1}\to\frac{1}{2}<1$, CONVERGES. $\sum\frac{\ln n}{n^{2}}$: a rational function with a logarithm, so comparison; bound it above by $\frac{1}{n^{3/2}}$ from some index on, CONVERGES. The test is chosen from the form of the term alone, before any computation.

Since the function $x\mapsto\frac{1}{x^{2}}$ is positive and decreasing, comparing the series with an integral gives $\int_{N+1}^{\infty}\frac{dx}{x^{2}}\leq R_{N}\leq\int_{N}^{\infty}\frac{dx}{x^{2}}$, that is $\dfrac{1}{N+1}\leq R_{N}\leq\dfrac{1}{N}$.

$S_{100}\approx 1.6349839$. The bounds give $\dfrac{1}{101}\approx 0.0099010\leq R_{100}\leq\dfrac{1}{100}=0.01$. The exact value is $\frac{\pi^{2}}{6}\approx 1.6449341$, so $R_{100}\approx 0.0099502$: it does fall between the bounds, close to their midpoint.

We want $R_{N}<10^{-3}$, and it is enough to ensure $\dfrac{1}{N}<10^{-3}$, that is $N>1000$. About a THOUSAND terms are therefore needed. The convergence is real but very slow, which is typical of p-series close to the boundary.

For the alternating series, the remainder is bounded by the first omitted term: $|R_{N}|\leq\dfrac{1}{(N+1)^{2}}$. We want $(N+1)^{2}>1000$, so $N+1>31.6$ and $N=31$ is enough. Thirty-one terms against a thousand: a factor of about thirty.

Because the partial sums of an alternating series with decreasing terms BRACKET the sum: the partial sums of even and odd index approach it from below and from above, closing in. The sum therefore always lies between two consecutive partial sums, and the gap between them is exactly the next term. In a series of positive terms, on the contrary, all the partial sums lie on the same side of the limit and the remainder adds up every omitted term instead of letting them cancel: here it is of the order of one over N, while the alternating remainder is of the order of one over N squared. The integrals bound that remainder very tightly; it is the remainder itself that is large.

Since $\alpha>0$, $\frac{1}{n^{\alpha}}\to 0$, and $\ln(1+u)\sim u$ gives $\left|a_{n}\right|=\ln\left(1+\frac{1}{n^{\alpha}}\right)\sim\frac{1}{n^{\alpha}}$.

Three checks. Alternating sign: the factor $(-1)^{n}$ provides it, and $\ln\left(1+\frac{1}{n^{\alpha}}\right)>0$ since the argument is greater than 1. Decrease: $n\mapsto\frac{1}{n^{\alpha}}$ decreases for $\alpha>0$, and the logarithm is increasing, so the composition decreases. Zero limit: $\ln(1+0)=0$. The three hypotheses hold for every $\alpha>0$, so $\mathcal{S}(\alpha)$ always converges. Note that convergence holds even for very small $\alpha$, where the decrease is extremely slow.

Absolute convergence amounts to the convergence of $\sum\ln\left(1+\frac{1}{n^{\alpha}}\right)$, which by the equivalent in a behaves like $\sum\frac{1}{n^{\alpha}}$: it converges if and only if $\alpha>1$. So $\mathcal{S}(\alpha)$ is absolutely convergent for $\alpha>1$, and CONDITIONALLY CONVERGENT for $0<\alpha\leq 1$.

For $\alpha=1$, $\ln\left(1+\frac{1}{n}\right)=\ln(n+1)-\ln n$. The partial sum of the absolute values telescopes: $\sum_{n=1}^{N}\ln\left(1+\frac{1}{n}\right)=\ln(N+1)$, which tends to $+\infty$. The divergence is thus confirmed directly, without going through the equivalent, and you get the rate as a bonus: the partial sum grows like $\ln N$. This is the boundary case of question c.

For $\alpha=2$, the remainder satisfies $\left|R_{N}\right|\leq\left|a_{N+1}\right|=\ln\left(1+\frac{1}{(N+1)^{2}}\right)\leq\dfrac{1}{(N+1)^{2}}$, using $\ln(1+u)\leq u$. It is therefore enough that $(N+1)^{2}>10^{4}$, that is $N+1>100$, and $N=100$ terms. The bound $\ln(1+u)\leq u$ is handy here: it avoids having to invert a logarithm.